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Row sums of triangle A131414.

Equals binomial transform of (1,5,3,0,0,0,...). Equals A051340 * (1,2,3,...).

a(n) is essentially the case -1 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} (k-2)*i-(k-3). Thus P_{-1}(n) = n*(5-3*n)/2 and a(n) = -P_{-1}(n+2). - Peter Luschny, Jul 08 2011

Beginning with n=2, a(n) is the falling diagonal starting with T(1,3) in A049777 (as a square array). - Bob Selcoe, Oct 27 2014

Maximum dimension of Euclidean spaces which suffice for every smooth compact Riemannian n-manifold to be realizable as a sub-manifold. - comment edited by Gene Ward Smith, Jan 15 2017

These numbers cannot be perfect squares. See the Hilliard link for a proof.

2nd elementary symmetric polynomial of n, n + 1 and n + 2: n(n+1) + n(n+2) + (n+1)(n+2). - Zak Seidov, Mar 23 2005

This sequence equals for n >= 2 the third right hand column of triangle A165674. Its recurrence relation leads to Pascal's triangle A007318. Crowley's formula for A080663(n-1) leads to Wiggen's triangle A028421 and the o.g.f. of this sequence, without the first term, leads to Wood's polynomials A126671. See also A165676, A165677, A165678 and A165679. - Johannes W. Meijer, Oct 16 2009

The Diophantine equation x(x+1) + (x+2)(x+3) = (x+y)^2 + (x-y)^2 has solutions x = a(n), y = 3n. - Bruno Berselli, Mar 29 2013

A simpler proof that these numbers can't be perfect squares can easily be constructed using congruences: If the equation x^2 = 3y^2 - 1 has a solution in positive integers, then x^2 = 2 mod 3. Obviously we can't have x = 0 mod 3, and x = 1 mod 3 doesn't work either because then x^2 = 1 mod 3 also. That leaves x = 2 mod 3, but then x^2 = 1 mod 3. - Alonso del Arte, Oct 19 2013

2*a(n+1) is surface area of a rectangular prism with consecutive integer sides: n, n+1, and n+2, (n>0). - Wesley Ivan Hurt, Sep 06 2014

Numbers m such that 3*m+3 is a square. So, these are the numbers m such that the system of equations x=sqrt(m-2yz), y=sqrt(m+1-2xz), z=sqrt(m+2-2xy) admits 3 real positive solutions whose sum is an integer. See the Rechtman link. - Michel Marcus, Jun 06 2020

As a rectangle, the accumulation array of A051340.

From Clark Kimberling, Feb 05 2011: (Start)

Here all the weights are divided by two where they aren't in Cahn.

As a rectangle, A141419 is in the accumulation chain

... < A051340 < A141419 < A185874 < A185875 < A185876 < ...

(See A144112 for the definition of accumulation array.)

row 1: A000027

col 1: A000217

diag (1,5,...): A000326 (pentagonal numbers)

diag (2,7,...): A005449 (second pentagonal numbers)

diag (3,9,...): A045943 (triangular matchstick numbers)

diag (4,11,...): A115067

diag (5,13,...): A140090

diag (6,15,...): A140091

diag (7,17,...): A059845

diag (8,19,...): A140672

(End)

Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011

Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x) (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N](i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012

Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013

n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014

T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014

A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016

Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018

From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)

A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).

The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.

Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.

(End)

Triangle read by rows, T(n,k) = A000217(n) - A000217(k), 0 <= k < n. - Philippe Deléham, Mar 07 2013

Subtriangle of triangle in A049780. - Philippe Deléham, Mar 07 2013

No primes and all composite numbers (except 2^x) are generated after the first two columns of the square array for this sequence. In other words, no primes and all composites except 2^x are generated when m-n >= 2. - Bob Selcoe, Jun 18 2013

Diagonal sums in the square array equal partial sums of squares (A000330). - Bob Selcoe, Feb 14 2014

From Bob Selcoe, Oct 27 2014: (Start)

The following apply to the triangle as a square array read by rows unless otherwise specified (see Table link);

Conjecture: There is at least one prime in interval [T(n,k), T(n,k+1)]. Since T(n,k+1)/T(n,k) decreases to (k+1)/k as n increases, this is true for k=1 ("Bertrand's Postulate", first proved by P. Chebyshev), k=2 (proved by El Bachraoui) and k=3 (proved by Loo).

Starting with T(1,1), The falling diagonal of the first 2 numbers in each column (read by column) are the generalized pentagonal numbers (A001318). That is, the coefficients of T(1,1), T(2,1), T(2,2), T(3,2), T(3,3), T(4,3), T(4,4) etc. are the generalized pentagonal numbers. These are A000326 and A005449 (Pentagonal and Second pentagonal numbers: n*(3*n+1)/2, respectively), interweaved.

Let D(n,k) denote falling diagonals starting with T(n,k):

Treating n as constant: pentagonal numbers of the form n*k + 3*k*(k-1)/2 are D(n,1); sequences A000326, 005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542 are formed by n = 1 through 12, respectively.

Treating k as constant: D(1,k) are (3*n^2 + (4k-5)*n + (k-1)*(k-2))/2. When k = 2(mod3), D(1,k), is same as D(k+1,1) omitting the first (k-2)/3 numbers in the sequences. So D(1,2) is same as D(3,1); D(1,5) is same as D(6,1) omitting the 6; D(1,8) is same as D(9,1) omitting the 9 and 21; etc.

D(1,3) and D(1,4) are sequences A095794 and A140229, respectively.

(End)

a(n) equals the number of vertices of the A256666(n)-th graph (see Illustration of initial terms in A256666 Links). - Ivan N. Ianakiev, Apr 20 2015